Question: Solve for $x$ and $y$ using substitution. ${-2x-4y = 0}$ ${y = -2x-3}$
Since $y$ has already been solved for, substitute $-2x-3$ for $y$ in the first equation. ${-2x - 4}{(-2x-3)}{= 0}$ Simplify and solve for $x$ $-2x+8x + 12 = 0$ $6x+12 = 0$ $6x+12{-12} = 0{-12}$ $6x = -12$ $\dfrac{6x}{{6}} = \dfrac{-12}{{6}}$ ${x = -2}$ Now that you know ${x = -2}$ , plug it back into $\thinspace {y = -2x-3}\thinspace$ to find $y$ ${y = -2}{(-2)}{ - 3}$ $y = 4 - 3$ $y = 1$ You can also plug ${x = -2}$ into $\thinspace {-2x-4y = 0}\thinspace$ and get the same answer for $y$ : ${-2}{(-2)}{ - 4y = 0}$ ${y = 1}$